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ajinkya bhole Grade: 10
        

in triangle ABC , D is the midpoint of AB . Pt.E is on AC such that AE=2EC . BE intersects CD at pt.F


Prove that BE=4EF

7 years ago

Answers : (1)

Askiitians Expert Sahil Arora - IITD
19 Points
										

Dear ajinkya


Let A be (0,0),  B(x1,y1),  C(x2,y2)


then  By section formula 


D=(x1/2,y1/2)


E=(2x2/3,2y2/3)


Let F(x3,y3)


let BF:FE=k:1    and DF:FC=m:1


then by section formula on line BE and CD we get


on BE........x3=(k*(2x2/3)+x1)/(k+1)  and y3= (k*(2y2/3)+y1)/(k+1)           ......(1)


On CD.......x3=(m*x2+x1/2)/(m+1)  and y3=(m*y2+y1/2)/(m+1)                        .........(2)


from (1) and (2)


(k*(2x2/3)+x1)/(k+1) =(m*x2+x1/2)/(m+1)        and     (k*(2y2/3)+y1)/(k+1) =(m*y2+y1/2)/(m+1) 


Compare to get coefficients of x1 and x2 or y1 and y2


we get


2k/3(k+1)=m/(m+1)


and 1/(k+1)=1/2(m+1)


solve these two eq. to get k=3;


So BF:FE=3:1


==>BE:FE=4:1


==>BE=4EF


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All the best ajinkya !!!




Regards,

Askiitians Experts

Sahil Arora
IIT Delhi
7 years ago
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